Finite dimensionality of $V_1times V_2times dotstimes V_m$ implies $V_j$ is finite dimensional $forall j$.
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Is my proof of the following proposition correct?
Proposition. Given that $V_1,V_2,dots,V_m$ are vector spaces such that $V_1times V_2timescdotstimes V_m$ is finite dimensional. Prove that $V_j$ is finite dimensional for each $j = 1,2,dots,m$.
Proof. Let $(u_11,u_12,dots,u_1m),(u_21,u_22,dots,u_2m),dots,(u_m1,u_m2,dots,u_mm)$
be the basis of $V_1times V_2timescdotstimes V_m$. Now let
$jin1,2,dots,m$, we show that $V_j =
operatornamespan(u_1j,u_2j,dots,u_mj)$.
Assume that $win V_j$, consequently $(0,0,dots,w_j,dots,0)in
V_1times V_2timescdotstimes V_m$ and thus for some
$lambda_1,lambda_2,dots,lambda_minmathbfF$ we have
$$(0,0,dots,w_j,dots,0)=sum_i=1^mlambda_i(u_i1,u_i2,dots,u_im)
= left(sum_i=1^mlambda_iu_i1,sum_i=1^mlambda_iu_i2,dots,sum_i=1^mlambda_iu_imright)$$
implying $$w_j = sum_i=1^mlambda_iu_ij$$ and thus
$w_jinoperatornamespan(u_1j,u_2j,dots,u_mj)$ and by
extension
$V_jsubseteqoperatornamespan(u_1j,u_2j,dots,u_mj)$, and
since $u_1j,u_2j,dots,u_mjin V_j$ it immediately follows that
$operatornamespan(u_1j,u_2j,dots,u_mj)subseteq V$.
$blacksquare$
linear-algebra proof-verification quotient-spaces
asked Sep 3 at 9:44
Atif Farooq
2,7762824
 |Â
show 2 more comments
up vote
0
down vote
favorite
Is my proof of the following proposition correct?
Proposition. Given that $V_1,V_2,dots,V_m$ are vector spaces such that $V_1times V_2timescdotstimes V_m$ is finite dimensional. Prove that $V_j$ is finite dimensional for each $j = 1,2,dots,m$.
Proof. Let $(u_11,u_12,dots,u_1m),(u_21,u_22,dots,u_2m),dots,(u_m1,u_m2,dots,u_mm)$
be the basis of $V_1times V_2timescdotstimes V_m$. Now let
$jin1,2,dots,m$, we show that $V_j =
operatornamespan(u_1j,u_2j,dots,u_mj)$.
Assume that $win V_j$, consequently $(0,0,dots,w_j,dots,0)in
V_1times V_2timescdotstimes V_m$ and thus for some
$lambda_1,lambda_2,dots,lambda_minmathbfF$ we have
$$(0,0,dots,w_j,dots,0)=sum_i=1^mlambda_i(u_i1,u_i2,dots,u_im)
= left(sum_i=1^mlambda_iu_i1,sum_i=1^mlambda_iu_i2,dots,sum_i=1^mlambda_iu_imright)$$
implying $$w_j = sum_i=1^mlambda_iu_ij$$ and thus
$w_jinoperatornamespan(u_1j,u_2j,dots,u_mj)$ and by
extension
$V_jsubseteqoperatornamespan(u_1j,u_2j,dots,u_mj)$, and
since $u_1j,u_2j,dots,u_mjin V_j$ it immediately follows that
$operatornamespan(u_1j,u_2j,dots,u_mj)subseteq V$.
$blacksquare$
linear-algebra proof-verification quotient-spaces
asked Sep 3 at 9:44
Atif Farooq
2,7762824
Yes, it is correct. There's just a subscript $j$ missing from the final $V$.
– Roberto Rastapopoulos
Sep 3 at 9:46
1
A better proof: $v_j to (0,0,...,0,v_j,0...,0)$ is a one-to-one linear map from $V_j$ into the product, so $V_j$ is finite dimensional.
– Kavi Rama Murthy
Sep 3 at 9:49
In the first line of your proof, you seem to assume that the dimension of $V_1times V_m$ is $m$. Other than that, the proof seems correct. I like the idea of @KaviRamaMurthy
– Thibaut Dumont
Sep 3 at 9:51
4
One little remark : you give a (not the) basis of $V_1times cdots times V_m$ that has $m$ vectors. Why would $V_1times cdots times V_m$ be $m$-dimensional? It's better to use different letters for different variables.
– Arnaud D.
Sep 3 at 9:51
Related : math.stackexchange.com/questions/1487489/…
– Arnaud D.
Sep 3 at 9:52
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is my proof of the following proposition correct?
Proposition. Given that $V_1,V_2,dots,V_m$ are vector spaces such that $V_1times V_2timescdotstimes V_m$ is finite dimensional. Prove that $V_j$ is finite dimensional for each $j = 1,2,dots,m$.
Proof. Let $(u_11,u_12,dots,u_1m),(u_21,u_22,dots,u_2m),dots,(u_m1,u_m2,dots,u_mm)$
be the basis of $V_1times V_2timescdotstimes V_m$. Now let
$jin1,2,dots,m$, we show that $V_j =
operatornamespan(u_1j,u_2j,dots,u_mj)$.
Assume that $win V_j$, consequently $(0,0,dots,w_j,dots,0)in
V_1times V_2timescdotstimes V_m$ and thus for some
$lambda_1,lambda_2,dots,lambda_minmathbfF$ we have
$$(0,0,dots,w_j,dots,0)=sum_i=1^mlambda_i(u_i1,u_i2,dots,u_im)
= left(sum_i=1^mlambda_iu_i1,sum_i=1^mlambda_iu_i2,dots,sum_i=1^mlambda_iu_imright)$$
implying $$w_j = sum_i=1^mlambda_iu_ij$$ and thus
$w_jinoperatornamespan(u_1j,u_2j,dots,u_mj)$ and by
extension
$V_jsubseteqoperatornamespan(u_1j,u_2j,dots,u_mj)$, and
since $u_1j,u_2j,dots,u_mjin V_j$ it immediately follows that
$operatornamespan(u_1j,u_2j,dots,u_mj)subseteq V$.
$blacksquare$
linear-algebra proof-verification quotient-spaces
asked Sep 3 at 9:44
Atif Farooq
2,7762824
Is my proof of the following proposition correct?
Proposition. Given that $V_1,V_2,dots,V_m$ are vector spaces such that $V_1times V_2timescdotstimes V_m$ is finite dimensional. Prove that $V_j$ is finite dimensional for each $j = 1,2,dots,m$.
Proof. Let $(u_11,u_12,dots,u_1m),(u_21,u_22,dots,u_2m),dots,(u_m1,u_m2,dots,u_mm)$
be the basis of $V_1times V_2timescdotstimes V_m$. Now let
$jin1,2,dots,m$, we show that $V_j =
operatornamespan(u_1j,u_2j,dots,u_mj)$.
Assume that $win V_j$, consequently $(0,0,dots,w_j,dots,0)in
V_1times V_2timescdotstimes V_m$ and thus for some
$lambda_1,lambda_2,dots,lambda_minmathbfF$ we have
$$(0,0,dots,w_j,dots,0)=sum_i=1^mlambda_i(u_i1,u_i2,dots,u_im)
= left(sum_i=1^mlambda_iu_i1,sum_i=1^mlambda_iu_i2,dots,sum_i=1^mlambda_iu_imright)$$
implying $$w_j = sum_i=1^mlambda_iu_ij$$ and thus
$w_jinoperatornamespan(u_1j,u_2j,dots,u_mj)$ and by
extension
$V_jsubseteqoperatornamespan(u_1j,u_2j,dots,u_mj)$, and
since $u_1j,u_2j,dots,u_mjin V_j$ it immediately follows that
$operatornamespan(u_1j,u_2j,dots,u_mj)subseteq V$.
$blacksquare$
linear-algebra proof-verification quotient-spaces
linear-algebra proof-verification quotient-spaces
asked Sep 3 at 9:44
Atif Farooq
2,7762824
asked Sep 3 at 9:44
Atif Farooq
2,7762824
asked Sep 3 at 9:44
Atif Farooq
2,7762824
asked Sep 3 at 9:44
Atif Farooq
2,7762824
asked Sep 3 at 9:44
Atif Farooq
2,7762824
2,7762824
Yes, it is correct. There's just a subscript $j$ missing from the final $V$.
– Roberto Rastapopoulos
Sep 3 at 9:46
1
A better proof: $v_j to (0,0,...,0,v_j,0...,0)$ is a one-to-one linear map from $V_j$ into the product, so $V_j$ is finite dimensional.
– Kavi Rama Murthy
Sep 3 at 9:49
In the first line of your proof, you seem to assume that the dimension of $V_1times V_m$ is $m$. Other than that, the proof seems correct. I like the idea of @KaviRamaMurthy
– Thibaut Dumont
Sep 3 at 9:51
4
One little remark : you give a (not the) basis of $V_1times cdots times V_m$ that has $m$ vectors. Why would $V_1times cdots times V_m$ be $m$-dimensional? It's better to use different letters for different variables.
– Arnaud D.
Sep 3 at 9:51
Related : math.stackexchange.com/questions/1487489/…
– Arnaud D.
Sep 3 at 9:52
 |Â
show 2 more comments
Yes, it is correct. There's just a subscript $j$ missing from the final $V$.
– Roberto Rastapopoulos
Sep 3 at 9:46
1
A better proof: $v_j to (0,0,...,0,v_j,0...,0)$ is a one-to-one linear map from $V_j$ into the product, so $V_j$ is finite dimensional.
– Kavi Rama Murthy
Sep 3 at 9:49
In the first line of your proof, you seem to assume that the dimension of $V_1times V_m$ is $m$. Other than that, the proof seems correct. I like the idea of @KaviRamaMurthy
– Thibaut Dumont
Sep 3 at 9:51
4
One little remark : you give a (not the) basis of $V_1times cdots times V_m$ that has $m$ vectors. Why would $V_1times cdots times V_m$ be $m$-dimensional? It's better to use different letters for different variables.
– Arnaud D.
Sep 3 at 9:51
Related : math.stackexchange.com/questions/1487489/…
– Arnaud D.
Sep 3 at 9:52
Yes, it is correct. There's just a subscript $j$ missing from the final $V$.
– Roberto Rastapopoulos
Sep 3 at 9:46
Yes, it is correct. There's just a subscript $j$ missing from the final $V$.
– Roberto Rastapopoulos
Sep 3 at 9:46
1
1
A better proof: $v_j to (0,0,...,0,v_j,0...,0)$ is a one-to-one linear map from $V_j$ into the product, so $V_j$ is finite dimensional.
– Kavi Rama Murthy
Sep 3 at 9:49
A better proof: $v_j to (0,0,...,0,v_j,0...,0)$ is a one-to-one linear map from $V_j$ into the product, so $V_j$ is finite dimensional.
– Kavi Rama Murthy
Sep 3 at 9:49
In the first line of your proof, you seem to assume that the dimension of $V_1times V_m$ is $m$. Other than that, the proof seems correct. I like the idea of @KaviRamaMurthy
– Thibaut Dumont
Sep 3 at 9:51
In the first line of your proof, you seem to assume that the dimension of $V_1times V_m$ is $m$. Other than that, the proof seems correct. I like the idea of @KaviRamaMurthy
– Thibaut Dumont
Sep 3 at 9:51
4
4
One little remark : you give a (not the) basis of $V_1times cdots times V_m$ that has $m$ vectors. Why would $V_1times cdots times V_m$ be $m$-dimensional? It's better to use different letters for different variables.
– Arnaud D.
Sep 3 at 9:51
One little remark : you give a (not the) basis of $V_1times cdots times V_m$ that has $m$ vectors. Why would $V_1times cdots times V_m$ be $m$-dimensional? It's better to use different letters for different variables.
– Arnaud D.
Sep 3 at 9:51
Related : math.stackexchange.com/questions/1487489/…
– Arnaud D.
Sep 3 at 9:52
Related : math.stackexchange.com/questions/1487489/…
– Arnaud D.
Sep 3 at 9:52
 |Â
show 2 more comments
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Yes, it is correct. There's just a subscript $j$ missing from the final $V$.
– Roberto Rastapopoulos
Sep 3 at 9:46
1
A better proof: $v_j to (0,0,...,0,v_j,0...,0)$ is a one-to-one linear map from $V_j$ into the product, so $V_j$ is finite dimensional.
– Kavi Rama Murthy
Sep 3 at 9:49
In the first line of your proof, you seem to assume that the dimension of $V_1times V_m$ is $m$. Other than that, the proof seems correct. I like the idea of @KaviRamaMurthy
– Thibaut Dumont
Sep 3 at 9:51
4
One little remark : you give a (not the) basis of $V_1times cdots times V_m$ that has $m$ vectors. Why would $V_1times cdots times V_m$ be $m$-dimensional? It's better to use different letters for different variables.
– Arnaud D.
Sep 3 at 9:51
Related : math.stackexchange.com/questions/1487489/…
– Arnaud D.
Sep 3 at 9:52